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Math Help - convergences subtruction question..

  1. #1
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    convergences subtruction question..

    i am given {An} , {Bn} so
    An->a
    Bn-An->0

    prove that Bn->a

    ??
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    i am given {An} , {Bn} so
    An->a
    Bn-An->0

    prove that Bn->a

    ??
    Is this in \mathbb{R}^n or in a generalized metric space? Ill assume its the former.

    We know that by the definition of a convergent sequence that if \xi_n\to\xi then \forall{\varepsilon>0} there exists N such that if N\leqslant{n} then d(\xi_n,\xi)<\varepsilon

    So we have that \forall\varepsilon>0 there exists a N such that N\leqslant{n} implies that d\left(A_n,a\right)<\frac{\varepsilon}{2}

    And that \forall\varepsilon>0 there exists a N_0 such that if N_0\leqslant{n} then d\left(B_n-A_n,0\right)<\frac{\varepsilon}{2}

    Now it should be pretty apparent that since the metric on \mathbb{R}^n is \left|\bold{x}-\bold{y}\right| that d\left(B_n-A_n,0\right)<\varepsilon\implies d\left(B_n,A_n\right)<\varepsilon


    So now consider

    d\left(B_n,a\right)<d\left(B_n,A_n\right)+d\left(A  _n,a\right)<\frac{\varepsilon}{2}+\frac{\varepsilo  n}{2}=\varepsilon

    For n>\max\left\{N,N_0\right\}
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  3. #3
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    i cant imagine where each group of numbers where e/2
    where N0?

    can you please make a line to describe on the axes
    where each term located
    ??
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    i cant imagine where each group of numbers where e/2
    where N0?

    can you please make a line to describe on the axes
    where each term located
    ??
    I'm sorry, I do not know what you mean? These are generic terms \varepsilon and \frac{\varepsilon}{2} are interchangable, they are both constants which you pick. You could have easily made both distances less than \varepsilon resulting in [ d\left(B_n,a\right)<2\varepsilon it gives the same result
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